# Full text of "Treatise On Analysis Vol-Ii"

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```188       XIII    INTEGRATION

(13.14.4)    With the hypotheses of(\ 3.14.3), the measure v is bounded if and only
if g is n-integrable; and v = 0 if and only if g is ^-negligible.

(13.14.5)    Let g^ g2 be two mappings ofX into R, such that gt is ^0 and
locally \jL-integrable. Then g2 is locally (g± - u)-integrable if and only if (with the
product convention 0/*(13.11)) g2g± is locally ju-integrable, and in that case we
have

By considering g2 and g2 , we reduce immediately to the case where
g2 J> 0. To say that g2 is locally (gl • //Hntegrable signifies (13.13.1) that, for
every /e JTR(X), the function g2f is (g± - XHntegrable, or equivalently
(13.14.3), that g2gifi\$ ju-integrable; which in turn means that g2gl is locally
/z-integrable. Also, if we put v = gl * ft and A = g2 • (gl • JLL), we have

= jf#2 dv =jfg20i dfJL by (13.14.3); hence the relation (13.14.5.1).

PROBLEMS

1.   Let X be a locally compact space, /x a positive measure on X, and (/„) a sequence of
functions belonging to  ^(X, /Lt). Write ftn=/n-^ and ju,,(A)=j   fndp, for any

/^.-measurable set A.

(a)    Show that if X is not compact it can happen that the sequence (fn) is unbounded
in L1, but that the sequence (//,„) is vaguely bounded.

(b)    Suppose that, for every subset A of X consisting of a single point, and for every
open subset A of X such that the measure induced by ^ on the frontier of A has finite
support, the sequence (/u.,,(A)) is bounded. Show that the sequence (/„) is bounded in
L1. (Show first that every point x0 eX has an open neighborhood U such that the
sequence of numbers |ftB|(U) is bounded. To do this, argue by contradiction, by showing
that otherwise it would be possible to define a strictly increasing sequence of integers
(nk\ a decreasing sequence (U*) of open neighborhoods of XQ , and a sequence (W*)
of /x-quadrable (Section 13.9, Problem 7) open sets, with the following properties:
tkcUfc-i,   KUfc-frro})^!/*,   lfi»<l(Ufc-{xo})^l   for   i<*.   W* c U*- Ck+1,
and finally

Consider the union W of the W*, and obtain a contradiction. Then show by a similar
argument that there exists a compact subset K of X such that the sequence
(W(X - K)) is bounded.)

(c)    Deduce from (b) that if the sequence (/u,n(A)) is bounded for every open set A in X,
then the sequence (fn) is bounded in L1.

(d)    On the interval [0, 1 ], take ftn to be the measure defined by a mass n at the point) when/^ 0. The
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